Let $\hat{i}, \hat{j}$ and $\hat{k}$ be the unit vectors along the three positive coordinate axes. Let \ $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$, \ $\vec{b} = \hat{i} + b_{2}\hat{j} + b_{3}\hat{k}$, $b_{2}, b_{3} \in \mathbb{R}$ \ $\vec{c} = c_{1}\hat{i} + c_{2}\hat{j} + c_{3}\hat{k}$, $c_{1}, c_{2}, c_{3} \in \mathbb{R}$ be three vectors such that $b_{2}b_{3} > 0$, $\vec{a} \cdot \vec{b} = 0$ and $\begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}\begin{pmatrix} 1 \\ b_{2} \\ b_{3} \end{pmatrix} = \begin{pmatrix} 3 - c_{1} \\ 1 - c_{2} \\ -1 - c_{3} \end{pmatrix}.$ Then, which of the following is/are TRUE ?

$\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ $\vec{b} = \hat{i} + b_{2}\hat{j} + b_{3}\hat{k}$ $\vec{c} = c_{1}\hat{i} + c_{2}\hat{j} + c_{3}\hat{k}$ $\text{Given: } b_{2}b_{3} > 0,\quad \vec{a} \cdot \vec{b} = 0$ $\begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}\begin{pmatrix} 1 \\ b_{2} \\ b_{3} \end{pmatrix} = \begin{pmatrix} 3 - c_{1} \\ 1 - c_{2} \\ -1 - c_{3} \end{pmatrix}$ $\text{To find: which of (1) - (4) is/are TRUE}$ $\text{Plan: } (1)\ \text{Matrix} \Rightarrow \vec{c}\times\vec{b} = \vec{a}-\vec{c}$ $(2)\ \text{dot } \vec{b} \Rightarrow \vec{b}\cdot\vec{c}=0\,;\quad (3)\ \text{dot } \vec{c} \Rightarrow \vec{a}\cdot\vec{c}=|\vec{c}|^2$ $(4)\ |\vec{b}|>\sqrt{10}\ \text{via } b_2b_3>0\,;\quad (5)\ |\vec{c}|\le|\vec{a}|=\sqrt{11}$ $\vec{a}\cdot\vec{b}=0 \Rightarrow \vec{a}\perp\vec{b}$ $[\vec{c}]_\times = \begin{pmatrix} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{pmatrix}$ $[\vec{c}]_\times\,\vec{b} = \vec{c}\times\vec{b}$ $(3-c_1,\ 1-c_2,\ -1-c_3) = \vec{a}-\vec{c}$ $\vec{c}\times\vec{b} = \vec{a}-\vec{c}$ $\vec{b}\cdot(\vec{c}\times\vec{b}) = \vec{b}\cdot(\vec{a}-\vec{c})$ $0 = \vec{b}\cdot\vec{a} - \vec{b}\cdot\vec{c}$ $\vec{b}\cdot\vec{a} = 0 \Rightarrow \vec{b}\cdot\vec{c} = 0$ $\vec{b}\cdot\vec{c}=0 \quad \checkmark\ \text{(option 2 is TRUE)}$ $\vec{c}\cdot(\vec{c}\times\vec{b}) = \vec{c}\cdot(\vec{a}-\vec{c})$ $0 = \vec{c}\cdot\vec{a} - \vec{c}\cdot\vec{c} = \vec{a}\cdot\vec{c} - |\vec{c}|^2$ $\vec{a}\cdot\vec{c} = |\vec{c}|^2$ $\vec{c}=0 \Rightarrow \vec{c}\times\vec{b}=0=\vec{a}-\vec{c}=\vec{a}\ \text{(false)}$ $\vec{a}\cdot\vec{c}=|\vec{c}|^2>0 \Rightarrow \text{option (1) is FALSE}$ $\vec{a}\cdot\vec{b} = 3(1) + 1(b_2) + (-1)(b_3) = 0$ $3 + b_2 - b_3 = 0$ $b_3 = b_2 + 3$ $b_2 b_3>0 \Rightarrow b_2(b_2+3)>0$ $b_2<-3 \ \text{ or } \ b_2>0$ $|\vec{b}|^2 = 1+b_2^2+b_3^2 = 1+b_2^2+(b_2+3)^2$ $|\vec{b}|^2 = 2b_2^2+6b_2+10$ $|\vec{b}|^2-10 = 2b_2(b_2+3) = 2b_2b_3$ $|\vec{b}|^2-10 = 2b_2b_3 > 0 \Rightarrow |\vec{b}|^2 > 10$ $|\vec{b}| > \sqrt{10} \quad \checkmark\ \text{(option 3 is TRUE)}$ $\vec{a}\cdot\vec{c} = |\vec{c}|^2 \le |\vec{a}|\,|\vec{c}|$ $|\vec{c}| \le |\vec{a}| = \sqrt{9+1+1} = \sqrt{11}$ $|\vec{c}| \le \sqrt{11} \quad \checkmark\ \text{(option 4 is TRUE)}$ $\vec{c}\ \text{lies inside the sphere of radius}\ |\vec{a}|=\sqrt{11}$ $\text{Answer: } (2),\ (3),\ (4)$ $\vec{b}\cdot\vec{c}=0\,;\quad |\vec{b}|>\sqrt{10}\,;\quad |\vec{c}|\le\sqrt{11}$